The Heaviside-Feynman Formula ( the 5fth equation in your question) is derived from the Liénard-Wiechert potentials,  which in turn are derived from the first four equations of yours. But from the Liénard-Wiechert potentials you could derive the following equation which is more convenient for our case :

\begin{equation}
\mathbf{E}(\mathbf{x},t) = \frac{q}{4\pi\epsilon_0}\left[\frac{\mathbf{n}-\boldsymbol{\beta}}{\gamma^2(1 - \boldsymbol{\beta}\cdot\mathbf{n})^3 R^2} \right]_{\mathrm{ret}} + \frac{q}{4\pi}\sqrt{\frac{\mu_0}{\epsilon_0}}\left[\frac{\mathbf{n}\times\left[(\mathbf{n}-\boldsymbol{\beta})\times \dot{\boldsymbol{\beta}}\right]}{(1 - \boldsymbol{\beta}\cdot\mathbf{n})^3 R}\right]_{\mathrm{ret}}
\tag{01}
\end{equation}
where

\begin{align}
\boldsymbol{\beta} & = \dfrac{\boldsymbol{\upsilon}}{c},\quad \beta=\dfrac{\upsilon}{c}, \quad \gamma= \left(1-\beta^{2}\right)^{-\frac12}
\tag{02a}\\
\dot{\boldsymbol{\beta}} & = \dfrac{\dot{\boldsymbol{\upsilon}}}{c}=\dfrac{\mathbf{a}}{c}
\tag{02b}\\
\mathbf{n} & = \dfrac{\mathbf{R}}{\Vert\mathbf{R}\Vert}=\dfrac{\mathbf{R}}{R}\equiv\dfrac{\mathbf{r'}}{r'}=\dfrac{\mathbf{r'}}{\Vert\mathbf{r'}\Vert}\equiv \mathbf{e}_{r'}
\tag{02c}
\end{align}
The electric field of equation (01) is given by equation 14.14 from Jackson's *'Classical Electrodynamics'*,3rd Edition.